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State the hybridisation of asterisked carbon in $\ce{CH3-CH=\overset{\ast}{C}=CH2}.$

I am not sure between $\mathrm{sp^2}$ and $\mathrm{sp}$ hybridization. Because of the presence of double bonds I think it should be $\mathrm{sp^2},$ but because of the two π-bonds I think $\mathrm{sp}$.

Which one is correct?

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  • $\begingroup$ The molecule you've drawn belongs to a class of molecules called "allenes." Here are some diagrams of the molecule's structure and bonding that may be helpful. $\endgroup$ – ron Oct 5 '14 at 12:39
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You must be clear that the hybridization is decided by the number of σ-electrons, so for $\ce{CH4}$ the number of σ-electrons is four and so its hybridization is $\mathrm{sp^3}.$

In case of $\ce{CH3-CH=\overset{\ast}{C}=CH2}$ the $\ce{\overset{\ast}{C}}$ is having two σ- and two π-bonds, so its hybridization is $\mathrm{sp}.$

While deciding hybridization we need to only care about bonds, and that too only the σ-bonds. Remember that possible hybridization for $\ce{C}$ are $\mathrm{sp^3},$ $\mathrm{sp^2}$ and $\mathrm{sp}.$

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As the asterisked carbon has two neighbors, the hybridization of this asterisked carbon is $\mathrm{sp},$ i.e. the angle $\ce{−CH=\overset{\ast}{C}=CH2}$ is 180°.

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I count the number of sigma bonds and the number of lone pairs, then add them, in this case we get only two sigma bonds and no lone pairs.

Now consider this as a reference table:

2=sp
3=sp2
4=sp3

So in this case it's sp.

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  • $\begingroup$ Welcome to Chem SE and thanks for your contribution! Don't get me wrong but I believe this doesn't add anything to the already accepted answer. If you write an answer to a question which already has an accepted answer than make sure that it adds some value to the overall topic being discussed. :) $\endgroup$ – Nisarg Bhavsar May 23 at 17:23

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